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 October 6th, 2012, 10:17 AM #1 Newbie   Joined: Jan 2012 Posts: 9 Thanks: 0 Irrational numbers How can I prove that sum of this twon numbers in this link is irrational Link- http://imageshack.us/photo/my-images/109/scanyr.jpg/ Thank you
 October 6th, 2012, 11:17 AM #2 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: Irrational numbers We know that $sqrt{2} \$ is not an integer, then you can use that result to show $\ 2^{\frac{1}{2}} + 3^{\frac{1}{3}} \$ is irrational by way of contradiction. Suppose $2^{\frac{1}{2}} + 3^{\frac{1}{3}}= \frac{a}{b}$ for integer $a,b \ and \ b \ne 0 \ and \ a \ne 0$ $3^{\frac{1}{3}}= \frac{a}{b} - sqrt{2}$ $$$3^{\frac{1}{3}}$$^3= \ $$\frac{a - b sqrt{2}}{b}$$^3$ $3b^3= $$a - b sqrt{2}$$^3$ Now look at this very carefully... a and b are integers. We know $b \ne 0 \$ because it was in the original denominator, we know $a \ne 0 \$ because the original sum is not zero. $\ 3b^3 \$ is an integer, $$$a - b sqrt{2}$$^3$ has no chance of being an integer because the expression inside the parenthesis is NOT an integer since $\ sqrt{2} \$ is NOT an integer. Conclusion: We assumed there were integers a,b to make the sum rational, we found this to be impossible therefore the sum is irrational. I'm sure others will post better proofs, wait for them.
October 7th, 2012, 07:23 PM   #3
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Re: Irrational numbers

Quote:
 $(a-b\sqrt{2})^3$has no chance of being an integer because the expression inside the parenthesis is NOT an integer since $\sqrt{2}$ is NOT an integer.
I am not sure I agree with this. $3^{\frac{1}{3}}$ is not an integer but ${(3^{\frac{1}{3}})}^3$ is an integer

 October 7th, 2012, 08:46 PM #4 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: Irrational numbers That's a good point Dougy, i'm going to have to rethink on it. If this 'thing'... $(a - b sqrt{2})$ ... is the cube root of an integer then my argument is useless. Maybe i can fix my argument? $(a - b sqrt{2})^3= a^3 - 3 sqrt{2}a^2b +6ab^2 - 2 sqrt{2}b^3$ Those radicals are going to prevent the expanded expression on the right from being an integer since a, b can't be zero. There is another possibility... $-3 sqrt{2}a^2b - 2 sqrt{2}b^3= 0$ $-b sqrt{2} $$3a^2 + 2b^2$$= 0$ For non-zero b this has no real number solutions. So, have i fixed my argument? Let me restate my argument (and change the wording). For non-zero integer a and b , $(a - b sqrt{2})^3$ cannot be an integer.
 October 7th, 2012, 09:01 PM #5 Senior Member   Joined: Nov 2011 Posts: 595 Thanks: 16 Re: Irrational numbers Yes I think that (almost) finishes the proof now! However I think you did not treat the most general case. You want to prove that $(a-b\sqrt{2})^3$ can not be an integer/ Now you say that is equivalent to prove that $a^3-3\sqrt{2}a^2b+6ab^2-2\sqrt{2}b^3$ is not an integer. Since, as you said, $a^3+6ab^2$ is an integer, that means this is equivalent to show that $-3\sqrt{2}a^2b-2\sqrt{2}b^3$ can not be an integer. Now I don't understand why you put this thing to be equal to zero? It could be equal to any integer no? In any case, the proof is saved, because if we assume there exists an integer P such that $-3\sqrt{2}a^2b-2\sqrt{2}b^3=P$ means $-b\sqrt{2}(3a^2+2b^2)=P$ means $\sqrt{2}=\frac{P}{-b(3a^2+2b^2)}$ which means that $\sqrt{2}$ is rational, which is absurd... edit: here P is a negative integerof course..so it is more like a relative number than an integer, question of wording!
 October 7th, 2012, 09:10 PM #6 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: Irrational numbers You are right again Dougy, in my mind i wanted to make the radicals 'go away' that's why i set it equal to zero. You are correct, I would have to rule out every integer. You saved the proof very nicely. I was not thorough...

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